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\begin{center}
\vskip 1cm{\LARGE\bf \LARGE\bf Hessenberg Matrices and Integer Sequences \\
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}
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\large {
Milan Janji\'c \\
Department of Mathematics and Informatics\\
University of Banja Luka \\
Republic of Srpska \\
Bosnia and Herzegovina \\
\href{mailto:agnus@blic.net}{\tt agnus@blic.net}\\
}
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\centerline{\emph{Dedicated to the memory of Professor Veselin Peri\'c} }
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\begin{abstract}
We consider a particular case of upper Hessenberg matrices, in which
all subdiagonal elements are $-1.$ We investigate three type of
matrices related to polynomials, generalized Fibonacci numbers, and
special compositions of natural numbers. We give the combinatorial
meaning of the coefficients of the characteristic polynomials of these
matrices.
\end{abstract}
\section{Introduction}
We investigate a particular case of upper Hessenberg matrices, in which all subdiagonal elements are $-1.$
Several mathematical objects may be represented as determinants of such matrices.
We consider three type of matrices related to polynomials, generalized Fibonacci numbers,
and a special kind of composition of natural numbers.
Our objective is to find the combinatorial meaning of the coefficients of the characteristic polynomials.
The coefficients of the characteristic polynomials of matrices of the
first type, that is, the sums of principal minors, are related to some
binomial identities.
The characteristic polynomials of matrices of the second kind give, as
a particular case, the so-called
convolved Fibonacci numbers defined by Riordan \cite{riordan}.
Coefficients of the characteristic polynomial of matrices of the third
kind are connected with a special kind of composition of natural
numbers, in which there are two different types of ones. These were
introduced by Deutsch \cite{eme},
and studied by Grimaldi \cite{papa}.
We prove several formulas which generate a number of sequences in
Slonae's {\it Encyclopedia} \cite{slo}.
The following result about upper Hessenberg matrices, which may be
easily proved by induction, will be used in this paper.
\begin{theorem}\label{t1} Let the matrix $P_n$ be defined by
\begin{equation}\label{mat}P_n= \left[\begin{array}{llllll} p_{1,1}&p_{1,2}&p_{1,3}&\cdots&p_{1,n-1}&p_{1,n}\\
-1&p_{2,2}&p_{2,3}&\cdots&p_{2,n-1}&p_{2,n}\\ 0&-1&p_{3,3}&\cdots&p_{3,n-1}&p_{3,n}\\ \vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\
0&0&0&\cdots&p_{n-1,n-1}&p_{n-1,n}\\ 0&0&0&\cdots&-1&p_{n,n}\end{array}\right],\end{equation}
and let the sequence $a_1,a_2,\ldots$ be defined by
\begin{equation}\label{niz}a_{n+1}=\sum_{i=1}^np_{i,n}a_i,\;(n=1,2,\ldots).\end{equation}
Then \[a_{n+1}=a_1\det
P_n,\;(n=1,2,\ldots).\]
\end{theorem}
The Fibonacci numbers are defined by
$F_0 = 0, F_1 = 1,$ and $F_n =
F_{n-1} + F_{n-2}$ for $n \geq 2$
Also, we use the well-known fact that the coefficients of the
characteristic polynomial of a matrix are, up to the sign,
sums of principal minors of the matrix.
\section{Polynomials}
We start our investigations with Hessenbeg matrices whose determinants
are polynomials.
According to Theorem \ref{t1} we have
\begin{proposition} Let the matrix $P_{n+1}$ be defined by: \[P_{n+1}=\left[\begin{array}{rrrrrr}
1&p_{1}&p_{2}&\cdots&p_{n-1}&p_{n}\\
-1&x&0&\cdots&0&0\\
0&-1&x&\cdots&0&0\\
\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\
0&0&0&\cdots&x&0\\
0&0&0&\cdots&-1&x\end{array}\right].\] Then
\[\det P_{n+1}=x^n+p_1x^{n-1}+\cdots+p_n.\]
\end{proposition}
We let $S_{n+1-k}(x)$ denote the sum of all principal minors of order $n+1-k$ of $P_{n+1},$ where $k=0,\ldots,n,$ and $S_0=1.$
\begin{proposition}\label{pr1} The following formulas are valid
\begin{equation}\label{e1}S_{n+1-k}(x)={n\choose k-1}x^{n-k+1}+x^{n-k}\sum_{i=0}^{n-k}p_{i}{n-i\choose k}x^{-i},\;(k=0,1,2,\ldots,n).\end{equation}
\end{proposition}
\begin{proof}
For $k=0$ equation (\ref{e1}) means that $S_{n+1}=\det P_{n+1},$ which is clear.
If we delete the first row and the first column of $P_{n+1}$ we
obtain a lower triangular matrix of order $n$ with $x$'s
on the main diagonal. Hence, all principal minors of order $n-k$
obtained by deleting the first row and the first column of $P_{n+1}$
and another $k-1$ rows and columns with the same indices are equal
to $x^{n-k+1}$. There are $n\choose k-1$ such minors. This gives
the first term in equation (\ref{e1}).
We shall next calculate the minor $M_{n+1-k}$ obtained by deleting the rows and columns with indices $2\leq m_11,$ we obtain an upper triangular block matrix
where the upper block is $P_{m-1}$ and thus its determinant is equal $\sum_{i=0}^{m-2}p_ix^{m-2-i}.$ The lower block is
a lower triangular matrix of order $n+1-m$ with $x$'s on the main diagonal.
It follows that $M_{n+1-k}(x)=\sum_{i=0}^{m-2}p_ix^{m-2-i}x^{n-k-m+2}.$ For a fixed $m>1$ there are ${n+1-m\choose k-1}$ such minors.
We thus obtain
\[S_{n+1-k}(x)={n\choose k-1}x^{n-k+1}+\sum_{m=2}^{n-k+2}\sum_{i=0}^{m-2}p_i{n-m+1\choose k-1}x^{n-k-i},\;(k=1,2,\ldots,n).\]
Changing the order of summation yields
\[S_{n+1-k}(x)={n\choose k-1}x^{n-k+1}+\sum_{i=0}^{n-k}p_ix^{n-k-i}\sum_{m=i+2}^{n-k+2}{n-m+1\choose k-1},\;(k=1,2,\ldots,n).\]
The corollary now holds by the following recurrence relation for binomial coefficients:
\[\sum_{i=0}^{n-k}{n-i\choose k}={n+1\choose k+1}.\]
\end{proof}
\begin{corollary} If
$x=1,\;p_i=1,\;(i=0,1,\ldots,n)$ then
\[S_{n+1}=n+1,\;S_{n+1-k}={n\choose k-1}+{n+1\choose k+1},\;(k=1,2,\ldots,n).\]
\end{corollary}
Several sequences from Sloane's OIES \cite{slo} are generated by the preceding formulas. We state some of them:
\seqnum{A000124},
\seqnum{A000217},
\seqnum{A001105},
\seqnum{A001845},
\seqnum{A004006},
\seqnum{A005744},
\seqnum{A005893},
\seqnum{A006522},
\seqnum{A017281},
\seqnum{A027927},
\seqnum{A056220},
\seqnum{A057979},
\seqnum{A080855},
\seqnum{A080856},
\seqnum{A080857},
\seqnum{A105163},
\seqnum{A121555},
\seqnum{A168050}.
\begin{corollary}\label{co1}
If
$x\not=1,\;p_i=1,\;(i=1,\ldots,n)$ in (\ref{e1}) then
\[S_{n+1}(x)=\frac{x^{n+1}-1}{x-1},\;S_{n+1-k}(x)={n\choose k-1}x^{n-k+1}+\sum_{j=0}^{n-k}{k+j\choose k}x^j,\;(k=0,1,\ldots,n).\]
\end{corollary}
According to the well-known binomial identity
\[{k+j\choose j}=(-1)^j{-k-1\choose j},\] we see that the second term
\[\sum_{j=0}^{n-k}{k+j\choose k}x^j,\] in the preceding equation is the partial sum of the expansion of the function $\frac{1}{(1-x)^{k+1}}$ into powers of $x.$
The characteristic matrix $Q_{n+1}(t)$ of $P_{n+1}$ has the form
\[Q_{n+1}(t)=-\left[\begin{array}{rrrrrr}
1-t&p_{1}&p_{2}&\cdots&p_{n-1}&p_{n}\\
-1&x-t&0&\cdots&0&0\\
0&-1&x-t&\cdots&0&0\\
\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\
0&0&0&\cdots&x-t&0\\
0&0&0&\cdots&-1&x-t\end{array}\right].\]
This is also a Hessenberg matrix. If $f_{n+1}(t)$ is the characteristic polynomial of $P_{n+1}$ then
\[f_{n+1}(t)=\det(Q_{n+1}(t))=(-1)^{n+1}\left[\sum_{k=1}^{n}p_k(x-t)^{n-k}+(1-t)(x-t)^n\right].\]
We thus obtain
\begin{proposition} The following equation holds
\begin{equation}\label{kp}\sum_{k=0}^{n+1}(-1)^{n-k+1}S_{n-k+1}(x)t^k=(-1)^{n+1}\left[\sum_{k=1}^{n}p_k(x-t)^{n-k}+(1-t)(x-t)^n\right].\end{equation}
\end{proposition}
As a consequence we shall prove a curious binomial identity.
\begin{proposition} Let $m,n\;(0\leq m\leq n)$ be arbitrary integers. Then
\[\sum_{j=m}^n\sum_{k=0}^{n-j}(-1)^k{k+j\choose j}{j\choose m}=1.\]
\end{proposition}
\begin{proof}
Take $p_i=1,\;(i=0,\ldots,n),$ $t=1,$ and $x+1$ instead of $x$ in (\ref{kp}). The left side becomes
\[L=1+\sum_{k=0}^n(-1)^{n-k+1}\left[{n\choose k-1}(x+1)^{n-k+1}+\sum_{j=0}^{n-k}(-1)^{n-k+1}{k+j\choose j}(x+1)^j\right].\]
It is easily seen that
\[\sum_{k=0}^n(-1)^{n-k+1}{n\choose k-1}(x+1)^{n-k+1}=(-x)^n-1.\]
Therefore,
\[L=(-x)^n+\sum_{k=0}^n\sum_{j=0}^{n-k}\sum_{m=0}^j(-1)^{n-k+1}{k+j\choose j}{j\choose m}x^m.\]
Changing the order of summation we have
\[L=(-x)^n+\sum_{m=0}^n\left[\sum_{j=m}^{n}\sum_{k=0}^{n-j}(-1)^{n-k+1}{k+j\choose j}{j\choose m}\right]x^m.\]
The right side of (\ref{kp}) has the form
$R=(-1)^{n+1}(x^{n-1}+x^{n-2}+\cdots+1).$
In this way we obtain
\[\sum_{m=0}^n\left[\sum_{j=m}^{n}\sum_{k=0}^{n-j}(-1)^{n-k+1}{k+j\choose j}{j\choose m}\right]x^m=(-1)^{n+1}(x^n+x^{n-1}+\cdots+1).\]
The proposition follows by comparing coefficients of the same powers of $x$ in this equation.
\end{proof}
\section{Generalized Fibonacci Numbers}
Consider the sequence recursively given by $l_1=1,l_2=x,l_{n+1}=yl_{n-1}+xl_n,\;(n>2),$ and an upper Hessenberg matrix $L_n$ of order $n$ defined by: \[L_n=\left[\begin{array}{rrrrrr} x&y&0&\cdots&0&0\\ -1&x&y&\cdots&0&0\\
0&-1&x&\cdots&0&0\\ \vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\ 0&0&0&\cdots&x&y\\ 0&0&0&\cdots&-1&x\end{array}\right].\]
From Theorem \ref{t1} we obtain \[\det L_n=l_{n+1},\;(n=1,2,\ldots).\]
For $l_1=l_2=x=y=1$ we have $l_{n+1}=f_{n+1},$ where $f_{n+1}$ is the $n+1$th Fibonacci number. The numbers $l_n$ are usually called generalized Fibonacci numbers.
\begin{proposition}\label{pp} Let $S_{n-k},\;(k=0,1,\ldots,n-1)$ be the sum of all principal minors of $L_n$ of order $n-k.$ Then
\begin{equation}\label{snkk}S_{n-k}=\sum_{j_1+j_2+\cdots+j_{k+1}=n+1}l_{j_1}l_{j_2}\cdots l_{j_{k+1}},\end{equation}
where the sum is taken over $j_t\geq
1,\;(t=1,2,\ldots,k+1).$ \end{proposition}
\begin{proof}
Principal minors of $L_n$ are some convolutions of $l_n.$ For example, the minor obtained by deleting the $i$th row and
the $i$th column of $L_n$ is obviously equal to $l_{i}\cdot l_{n-i+1}.$ Therefore, the principal minor $M(i_1,i_2,\ldots,i_k)$ of $L_n$
obtained by deleting the rows and columns with indices $1\leq i_12).\]
It follows from Identity 2 in \cite{ben} that $a_{n+1}=f_{2n+1}.$
\end{proof}
\begin{proposition}\label{pp1} Let $S_{n-k},\;(0\leq k\leq n)$ be the sum of all minors of order $n-k$ of $G_n.$ Then
\[S_{n-k}=\sum_{j_1+j_2+\cdots+j_{k+1}=n-k}f_{2j_1+1}f_{2j_2+1}\cdots f_{2j_{k+1}+1},\] where the sum is taken over $j_t\geq
0,\;(t=1,2,\ldots,k+1).$ \end{proposition}
\begin{proof}
Let $M(i_1,i_2,\ldots,i_k)$ be the minor of order $n-k$ obtained by deleting the rows and columns with indices $1\leq i_1